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Mass of Hanging Block

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A 2kg block (m1) resting on a plane inclined 37 degrees is connected by a rope through a pulley to a block (m2) hanging free. The coefficient of static friction on m1 is 0.2. What is the mass of m2 if both masses are at rest? How about if both masses are moving at constant velocity?

    2. Relevant equations
    For m1:

    For m2:

    3. The attempt at a solution
    For m1:

    F=T-mgsinӨ=0, therefore T=mgsinӨ=11.81N

    Since mu=0.2, Fs=mu(N)=(0.2)(15.66N)=3.13N, therefore since Fs is less than T, the system will continue to accelerate upward.

    For m2:
    F=mg-T=0, therefore T=mg

    If both masses are at rest, the sum of y-direction forces on m2 must equal zero, therefore for m2, T-mg=0, thus:

    11.81N/9.81m/s/s = m
    m=1.2kg.......so in the end, does mu have any affect on the calculation of the mass for m2? Or does it and I've forgotten to calculate something?

    Also if both masses are moving at constant velocity (ie: no acceleration), the mass would be same as computed above.

    Thanks in advance for any tips.
  2. jcsd
  3. Mar 19, 2007 #2


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    There is a component of weight of m1 acting down the incline and there is the friction (static) component.

    Usually the coefficient of kinetic friction is less than that of static friction.
    Last edited: Mar 19, 2007
  4. Mar 19, 2007 #3


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    1.) The static friction force is not always u_s(N). Often, it is less.
    2.) The problem should have specified u_k for the constant velocity case. One can only assume that u_s =u_k. Even with that assumption, the results for m2 will not necessarily be the same for the 'at rest' and 'constant velocity' case.
  5. Mar 19, 2007 #4
    Sorry for the typo. u_k is 0.2. There is no specified u_s in the given problem. That being said, if f_k=u_k(f_N), where is the best point to incorporate this equation into the given work? It appears 1.2kg would be the mass if the system is frictionless in nature......but friction changes everything.
  6. Mar 19, 2007 #5


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    As Astronuc implied, you neglected the friction force in your application of Newton 1 for the constant velocity case. That will give you the solution to m2 for the constant velocity case. For the at rest case, friction is less than or equal to u_s(N), so you'd get a range of values for m2, starting with 1.2 Kg if you did the math correctly, up to some value depending on what u_s(N) is.
  7. Mar 20, 2007 #6
    Taking the advice of PhantomJay and Astronuc, let me redo part of the problem from scratch:

    Constant Velocity:

    The force pulling m1 downward is mgsinӨ, so

    The frictional force opposing this force is u_k(N) [where N is the Normal force, or in this case, mgcosӨ], so


    The weight of m2 is also opposing this force, so
    m2=Mg [where M is the mass value for m2]

    Since the block is moving with constant velocity the net force is zero, so
    Dividing by -1 eliminates the negative signs on both sides, so

    Does that part look right?
  8. Mar 21, 2007 #7


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    That looks right if the block m1 is moving at constant velocity down the plane. Now you must ask what the mass M must be if the block m1 is moving up the plane at constant velocity? And then,what does that tell you if the block m1 is at rest?
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